Problem 35 - Chemistry The Molecular Nature of Matter

Question

Barium sulfate is so insoluble that it can be swallowed without significant danger even though \(\mathrm{Ba}^{2+}\) is toxic. At \(25{ }^{\circ} \mathrm{C}, 1.00 \mathrm{~L}\) of water dissolves only \(0.00245 \mathrm{~g}\) of \(\mathrm{BaSO}_{4} .\) Calculate the molar solubility and \(K_{\mathrm{sp}}\) for \(\mathrm{BaSO}_{4}\).

Step-by-Step Solution

Answer

The molar solubility of \(\mathrm{BaSO}_{4}\) is \(1.049 \times 10^{-5}\) mol/L and the \(K_{\text{sp}}\) is \(1.10 \times 10^{-10}\).

Step 1: Calculate the molar solubility

Convert the mass of \(\mathrm{BaSO}_{4}\) to moles by using its molar mass (233.38 g/mol). Molar solubility is the number of moles of \(\mathrm{BaSO}_{4}\) that dissolve in 1 L of water: \[\text{Molar solubility} = \frac{0.00245\,\text{g}}{233.38\,\text{g/mol}}\]

Step 2: Write the dissociation equation

Write the balanced chemical equation for the dissociation of \(\mathrm{BaSO}_{4}\) in water: \[\mathrm{BaSO}_{4} (s) \leftrightarrow \mathrm{Ba}^{2+} (aq) + \mathrm{SO}_{4}^{2-} (aq)\]

Step 3: Express the solubility product constant \(K_{\text{sp}}\)

The solubility product constant \(K_{\text{sp}}\) is the product of the molar concentrations of the ions raised to the power of their stoichiometric coefficients: \[K_{\text{sp}} = [\mathrm{Ba}^{2+}][\mathrm{SO}_{4}^{2-}]\]

Step 4: Calculate the \(K_{\text{sp}}\)

Since mole ratio of \(\mathrm{Ba}^{2+}\) to \(\mathrm{BaSO}_{4}\) is 1:1 and the same for \(\mathrm{SO}_{4}^{2-}\), use the molar solubility found in Step 1 for the concentrations: \[K_{\text{sp}} = (\text{Molar solubility of } \mathrm{Ba}^{2+}) \times (\text{Molar solubility of } \mathrm{SO}_{4}^{2-})\]

Step 5: Finalize the calculation

Plug the molar solubility into the equation from Step 4 to find the \(K_{\text{sp}}\): \[K_{\text{sp}} = (\frac{0.00245}{233.38})^2\] Solve for \(K_{\text{sp}}\) to get the final value.

Key Concepts

Solubility Product Constant (Ksp)

The solubility product constant, abbreviated as Ksp, is a pivotal concept when discussing the solubility of sparingly soluble salts. Ksp provides us a quantitative measure of a salt's solubility in water at a given temperature. It is particularly important for predicting the extent of dissolution and whether a precipitate will form in a solution.

In technical terms, Ksp is defined for a salt that dissociates into its component ions. For example, consider a generic salt AB that dissociates into A+ and B-. The Ksp expression for this would be:
\[K_{\text{sp}} = [A^+][B^-]\]
Here, the brackets denote the molar concentration of the ions in moles per liter. It is important to note that solid AB does not appear in this expression because its concentration is constant as a pure solid.

A higher Ksp value indicates a more soluble compound, allowing us to compare the solubility of different salts under similar conditions. The solubility product is also essential when dealing with solubility equilibria, as it aids in understanding how factors like the common ion effect can shift equilibria and affect solubility.

Dissociation Equation in Chemistry

Dissociation in chemistry refers to the process wherein molecules or ionic compounds separate or split into smaller particles such as atoms, ions, or radicals, usually in a reversible manner. In the context of solubility, we focus on the dissociation of ionic compounds in solution.

The dissolution of an ionic compound can be represented through a dissociation equation. For instance, a compound like barium sulfate (BaSO4), when placed in water, dissociates into its respective ions as shown in the balanced chemical equation:
\[\mathrm{BaSO}_4 (s) \leftrightarrow \mathrm{Ba}^{2+} (aq) + \mathrm{SO}_4^{2-} (aq)\]
The 's' and 'aq' denote the solid state and aqueous solution, respectively. This equation illustrates how the solid salt dissociates into barium ions (Ba^2+) and sulfate ions (SO4^2-) when it dissolves. Understanding this equation is crucial for visualizing the process at the molecular level and for establishing the stoichiometric relationships necessary to determine solubility and calculate the solubility product constant.

Stoichiometry in Solubility

Stoichiometry in solubility involves the quantitative relationship between the amount of solute and the resulting ions in a solution. As salts dissolve in water, they disassociate into their constituent ions, and stoichiometry allows us to determine the number of moles of each ion produced from the salt. This is essential for calculating the solubility product constant (Ksp).

For a salt like barium sulfate (\(\mathrm{BaSO}_4\)), the stoichiometry is straightforward because each mole of \(\mathrm{BaSO}_4\) that dissolves yields one mole of \(\mathrm{Ba}^{2+}\) ions and one mole of \(\mathrm{SO}_4^{2-}\) ions. Using the molar mass of \(\mathrm{BaSO}_4\), one can convert the mass of \(\mathrm{BaSO}_4\) dissolved in water to moles, equal to the molar solubility. Subsequently, you can utilize these values to calculate the Ksp by multiplying the concentration of the barium ions by the concentration of the sulfate ions.

Remember, the stoichiometric coefficients in the balanced dissociation equation match the exponents on the ion concentrations in the Ksp expression. In more complex scenarios, where the dissociation is not a 1:1 ratio, stoichiometry becomes even more crucial to finding the relationship between the dissolved salt and the resulting ion concentrations.